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Let V and W be vector spaces over a field (or more generally, modules over a ring) and let T be a linear map from V to W. If 0W is the zero vector of W, then the kernel of T is the preimage of the zero subspace {0W}; that is, the subset of V consisting of all those elements of V that are mapped by T to the element 0W. The kernel is usually denoted as ker T, or some variation thereof:

Since a linear map preserves zero vectors, the zero vector 0V of V must belong to the kernel. The transformation T is injective if and only if its kernel is reduced to the zero subspace.

The kernel ker T is always a linear subspace of V. Thus, it makes sense to speak of the quotient spaceV/(ker T). The first isomorphism theorem for vector spaces states that this quotient space is naturally isomorphic to the image of T (which is a subspace of W). As a consequence, the dimension of V equals the dimension of the kernel plus the dimension of the image.

Let G and H be groups and let f be a group homomorphism from G to H.
If eH is the identity element of H, then the kernel of f is the preimage of the singleton set {eH}; that is, the subset of G consisting of all those elements of G that are mapped by f to the element eH.
The kernel is usually denoted ker f (or a variation).
In symbols:

Since a group homomorphism preserves identity elements, the identity element eG of G must belong to the kernel.
The homomorphism f is injective if and only if its kernel is only the singleton set {eG}. This is true because if the homomorphism f is not injective, then there exists a,b∈G{\displaystyle a,b\in G} with a≠b{\displaystyle a\neq b} such that f(a)=f(b){\displaystyle f(a)=f(b)}. This means that f(a)f(b)−1=eH{\displaystyle f(a)f(b)^{-1}=e_{H}}, which is equivalent to stating that f(ab−1)=eH{\displaystyle f(ab^{-1})=e_{H}} since group homomorphisms carry inverses into inverses and since f(a)f(b−1)=f(ab−1){\displaystyle f(a)f(b^{-1})=f(ab^{-1})}. In other words, ab−1∈ker⁡f{\displaystyle ab^{-1}\in \operatorname {ker} f}. Conversely, if there exists an element g≠eG∈ker⁡f{\displaystyle g\neq e_{G}\in \operatorname {ker} f}, then f(g)=f(eG)=eH{\displaystyle f(g)=f(e_{G})=e_{H}}, thus f is not injective.

Let R and S be rings (assumed unital) and let f be a ring homomorphism from R to S.
If 0S is the zero element of S, then the kernel of f is its kernel as linear map over the integers, or, equivalently, as additive groups. It is the preimage of the zero ideal {0S}, which is, the subset of R consisting of all those elements of R that are mapped by f to the element 0S.
The kernel is usually denoted ker f (or a variation).
In symbols:

Since a ring homomorphism preserves zero elements, the zero element 0R of R must belong to the kernel.
The homomorphism f is injective if and only if its kernel is only the singleton set {0R}.
This is always the case if R is a field, and S is not the zero ring.

Since ker f contains the multiplicative identity only when S is the zero ring, it turns out that the kernel is generally not a subring of R. The kernel is a subrng, and, more precisely, a two-sided ideal of R.
Thus, it makes sense to speak of the quotient ringR/(ker f).
The first isomorphism theorem for rings states that this quotient ring is naturally isomorphic to the image of f (which is a subring of S). (note that rings need not be unital for the kernel definition).

To some extent, this can be thought of as a special case of the situation for modules, since these are all bimodules over a ring R:

R itself;

any two-sided ideal of R (such as ker f);

any quotient ring of R (such as R/(ker f)); and

the codomain of any ring homomorphism whose domain is R (such as S, the codomain of f).

However, the isomorphism theorem gives a stronger result, because ring isomorphisms preserve multiplication while module isomorphisms (even between rings) in general do not.

Let M and N be monoids and let f be a monoid homomorphism from M to N.
Then the kernel of f is the subset of the direct productM × M consisting of all those ordered pairs of elements of M whose components are both mapped by f to the same element in N.
The kernel is usually denoted ker f (or a variation).
In symbols:

Since f is a function, the elements of the form (m,m) must belong to the kernel.
The homomorphism f is injective if and only if its kernel is only the diagonal set {(m,m) : m in M}.

It turns out that ker f is an equivalence relation on M, and in fact a congruence relation.
Thus, it makes sense to speak of the quotient monoidM/(ker f).
The first isomorphism theorem for monoids states that this quotient monoid is naturally isomorphic to the image of f (which is a submonoid of N),(for the congruence relation).

This is very different in flavour from the above examples.
In particular, the preimage of the identity element of N is not enough to determine the kernel of f.

Let A and B be algebraic structures of a given type and let f be a homomorphism of that type from A to B.
Then the kernel of f is the subset of the direct productA × A consisting of all those ordered pairs of elements of A whose components are both mapped by f to the same element in B.
The kernel is usually denoted ker f (or a variation).
In symbols:

Note that the definition of kernel here (as in the monoid example) doesn't depend on the algebraic structure; it is a purely set-theoretic concept.
For more on this general concept, outside of abstract algebra, see kernel of a function.

To be specific, let A and B be Mal'cev algebraic structures of a given type and let f be a homomorphism of that type from A to B. If eB is the neutral element of B, then the kernel of f is the preimage of the singleton set {eB}; that is, the subset of A consisting of all those elements of A that are mapped by f to the element eB.
The kernel is usually denoted ker f (or a variation). In symbols:

Since a Mal'cev algebra homomorphism preserves neutral elements, the identity element eA of A must belong to the kernel. The homomorphism f is injective if and only if its kernel is only the singleton set {eA}.

The notion of ideal generalises to any Mal'cev algebra (as linear subspace in the case of vector spaces, normal subgroup in the case of groups, two-sided ideals in the case of rings, and submodule in the case of modules).
It turns out that ker f is not a subalgebra of A, but it is an ideal.
Then it makes sense to speak of the quotient algebraG/(ker f).
The first isomorphism theorem for Mal'cev algebras states that this quotient algebra is naturally isomorphic to the image of f (which is a subalgebra of B).

The connection between this and the congruence relation for more general types of algebras is as follows.
First, the kernel-as-an-ideal is the equivalence class of the neutral element eA under the kernel-as-a-congruence. For the converse direction, we need the notion of quotient in the Mal'cev algebra (which is division on either side for groups and subtraction for vector spaces, modules, and rings).
Using this, elements a and b of A are equivalent under the kernel-as-a-congruence if and only if their quotient a/b is an element of the kernel-as-an-ideal.

Sometimes algebras are equipped with a nonalgebraic structure in addition to their algebraic operations.
For example, one may consider topological groups or topological vector spaces, with are equipped with a topology.
In this case, we would expect the homomorphism f to preserve this additional structure; in the topological examples, we would want f to be a continuous map.
The process may run into a snag with the quotient algebras, which may not be well-behaved.
In the topological examples, we can avoid problems by requiring that topological algebraic structures be Hausdorff (as is usually done); then the kernel (however it is constructed) will be a closed set and the quotient space will work fine (and also be Hausdorff).